Q. The sodium salt of a weak acid is hydrolysed to the extent of 3% in 0.1 M solution in water at $25^\circ C$ . If $K_{a}$ for weak acid is $1.3\times 10^{- 10}$ . The ionic product of water is

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A

$1.17\times 10^{- 14}$

0%

B

$1.17\times 10^{14}$

0%

C

$11.7\times 10^{- 14}$

100%

D

$11.7\times 10^{- 10}$

0%

Solution:

$K_{H}=\alpha ^{2}C$
$\left[_{\text{hydrolysis}}^{\text{α} \, \text{=} \, \text{degree} \, \text{of}}\right]$
$=\left(\right.0.03\left(\left.\right)^{2}\times 0.1=9\times 10^{- 5}$
$K_{H}=\frac{K_{w}}{K_{a}}$
or $K_{w}=K_{H}\times K_{a}$
$=9\times 10^{- 5}\times 1.3\times 10^{- 10}$
$=1.17\times 10^{- 14}$ .

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